Find a number field whose unit group is isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}$

Find a number field whose unit group is isomorphic to $$\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}.$$

I'm trying to use Dirichlet's Unit Theorem to solve this problem. It states that if $$K$$ is a number field of signature $$(r,s)$$ and $$\mu_K$$ is the set of roots of unity in $$K$$, then the unit group $$\mathcal{O}_K^{\times}$$ of the ring of integers is isomorphic to $$\mu_K \times \mathbb{Z}^{r+s−1}$$ as an abelian group. So I suppose I want $$r+s-1=1,$$ or $$r+s=2$$. This forces $$(r,s)=(0,2)$$ because if there's at least one real embedding then $$\mu_K$$ is just $$\{\pm 1\}$$ so not $$\mathbb{Z}/4\mathbb{Z}$$.

Therefore I need a number field of degree $$4$$ with four complex embeddings and whose set of roots of unity is isomorphic to $$\mathbb{Z}/4\mathbb{Z}$$. The cyclotomic field $$\mathbb{Q}(\zeta_5)$$ doesn't work because it has more than $$4$$ elements in its set of roots of unity (and $$\mathbb{Q}(\zeta_4)$$ doesn't work because it doesn't have degree $$4$$). I suppose it will have to be $$\mathbb{Q}(\alpha)$$ where the minimal polynomial of $$\alpha$$ has degree $$4$$ but I haven't been able to find an example. Hints would be appreciated.

• If the set of roots of unity in your $K$ has to be cyclic of order $4$ then it has to be $\{\pm1,\pm i\}$ under any complex embedding. Thus, I would search your $K$ among those of the form $\Bbb Q(i,\sqrt{d})$ with $d<0$ to start with. – Andrea Mori Mar 15 at 20:35
• So perhaps I could take $\mathbb{Q}(i, \sqrt{2})$? That has degree $4$ and four complex embeddings. Its set of roots of unity includes $\pm 1,\pm i$, so it would suffice to show there are no other roots of unity in this set. Any other root of unity would need to be a primitive $m$th root of unity with $m=3$ or $m \geq 5$. I'm not sure how to show this is impossible though. – AlephNull Mar 15 at 21:12
• Does $\Bbb Q(\zeta_8)$ get you where you want to be? – Robert Shore Mar 15 at 22:21
• @RobertShore I considered that but I'm not sure what made me refute it. It certainly has degree $4$ and four complex embeddings. But doesn't that have more than $4$ roots of unity? – AlephNull Mar 15 at 22:50
• @AlephNull I think your example of $\mathbb{Q}(i, \sqrt{2})$ works. You can check that the roots of unity in $\mathbb{Q}(\zeta_n)$ are $\mu_n$ if $n$ is even and $\mu_{2n}$ if $n$ is odd. (Intuitively, this is because multiplying by $-1$ will double the order of an odd root of unity) – vxnture Mar 15 at 23:00

You are almost there! Since $$K$$ has a torsion element of order $$4$$, it contains $$\zeta_4$$ and thus contains $$\mathbf{Q}(\zeta_4)$$. Dirichlet's unit theorem then says that $$K$$ must be degree $$4$$ and thus a quadratic extension of $$\mathbf{Q}(\zeta_4)$$.

Now suppose that $$K$$ is any quadratic extension of $$\mathbf{Q}(\zeta_4) = \mathbf{Q}(\sqrt{-1})$$. The signature of $$K$$ is $$(0,2)$$ and so $$K$$ has unit rank one. Also, $$K$$ has an element $$\zeta_4$$ of order $$4$$. The only thing that remains is to find a $$K$$ that doesn't have any extra torsion. But the torsion subgroup of a number field is always cyclic and generated by a $$m$$th root of unity, or a $$4n$$th root of unity in this case since we already have a $$4$$th root of unity. So you just have to ensure that

$$\mathbf{Q}(\zeta_{4n}) \not\subset K$$

for any $$n > 1$$. The degree of $$\mathbf{Q}(\zeta_{4n})$$ is (Euler's $$\varphi$$ function) $$\varphi(4n)$$. This is $$> 4$$ for $$n \ge 4$$. So the answer is:

$$K$$ can be any quadratic extension of $$\mathbf{Q}(\zeta_4)$$ which doesn't equal $$\mathbf{Q}(\zeta_8)$$ or $$\mathbf{Q}(\zeta_{12})$$.

Since $$\mathbf{Q}(\zeta_8) = \mathbf{Q}(\sqrt{-1},\sqrt{2}) = \mathbf{Q}(\sqrt{-1},\sqrt{-2})$$ and $$\mathbf{Q}(\zeta_{12}) = \mathbf{Q}(\sqrt{-1},\sqrt{-3}) = \mathbf{Q}(\sqrt{-1},\sqrt{2})$$, you can find many such $$K$$, for example $$K = \mathbf{Q}(\sqrt{-1},\sqrt{d})$$ for any squarefree $$\pm d > 3$$. These are not the only examples --- the others are precisely all the quadratic extensions $$K/\mathbf{Q}(\sqrt{-1})$$ which are non-Galois over $$\mathbf{Q}$$ such as $$\mathbf{Q}(i,\sqrt{3 + 4 i})$$.

• Perfect answer, and this only uses results proved in my course. – AlephNull Mar 18 at 19:21